Let $n> 1$ and $k_i\in\mathbb{N}_0$, $1\leq i\leq n$. What is the value of

$$\int_0^1dx_1\int_0^{1-x_1}dx_2\int_0^{1-(x_1 + x_2)}dx_3 \cdots \int_0^{1 - \sum_{i = 1}^{n -1}x_i} \binom{N}{k_1\, k_2\cdots k_n}\prod_{i = 1}^n x_i^{k_i} dx_n\text{,}$$ where $N = \sum_i k_i$?

The reason why I need to compute the integral is that I want to find $\int_\vec{p} P(\vec{k}\mid \vec{p})d\vec{p}$, where $\vec{k}$ follows the multinomial distribution, when $\vec{p}$ and $N = \sum_i k_i$ are given: $$P(\vec{k}\mid \vec{p}) = \binom{N}{k_1\, k_2\cdots k_n}\prod_i p_i^{k_i}\text{.}$$

For $n = 2$, the result is $\binom{N}{k_1\; k_2}B(k_1 + 1, k_2+2)$, where $B$ is Beta function, which can be simplified to $1/(N + 1)( N +2)$. However, I cannot solve the problem in general case.


See Gradsteyn&Ryzhik, 4.634.

With the notation $D=\left\{0\le x_i\le 1,0\le i\le n, x_1+x_2+\ldots+x_n\le 1\right\}$ one has \begin{align} &\int_0^1dx_1\int_0^{1-x_1}dx_2\int_0^{1-(x_1 + x_2)}dx_3 \cdots \int_0^{1 - \sum_{i = 1}^{n -1}x_i} \prod_{i = 1}^n x_i^{k_i} dx_n=\\ &=\int\limits_D \prod_{i = 1}^n x_i^{k_i}dx_i=\frac{\displaystyle\prod_{i=1}^n\Gamma(k_i+1)}{\Gamma(k_1+k_2+\ldots+k_n+n+1)}=\frac{\displaystyle\prod_{i=1}^nk_i!}{(N+n)!} \end{align}

  • $\begingroup$ I have found the book, there is a solution for an even more general integral. However, the answer in the book and yours also, lack any explanation. Can you explain, how do you get from $\int_D \prod x_i^{k_i} dx_i$ to the result? $\endgroup$ – Antoine May 19 '16 at 23:10
  • $\begingroup$ @Antoine In the same book one can find reference to Fichtenholz's course of mathematical analysis, vol III, where this integral is calculated. This is a textbook material, no need to replicate it here. $\endgroup$ – Nemo May 20 '16 at 10:33
  • $\begingroup$ There are many ways to calculte it. e.g. by induction using linear change of variables. or deforming $D$ into n-dimensional sphere by change of variables where the integral will be a product of n one-dimensional integrals. $\endgroup$ – Nemo May 20 '16 at 12:46

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